Question

The coefficient of \(x^{-5}\) in the binomial expansion of \(\left(\frac{x+1}{x^{2/3} - x^{1/3} + 1} - \frac{x-1}{x - x^{1/2}}\right)^{10}\), where \(x \neq 0, 1\), is :

• A. 1
• B. 4
• C. \(-4\)
• D. \(-1\)

Hint:

Simplify each fraction using a^3 + 1 = (a + 1)(a^2 - a + 1) and b^2 - 1 = (b - 1)(b + 1).

Answer 1

The Correct Option is A

Let \(a = x^{1/3}\). Then \(\frac{x+1}{x^{2/3} - x^{1/3} + 1} = \frac{a^3+1}{a^2 - a + 1} = a + 1 = x^{1/3} + 1\).
Let \(b = x^{1/2}\). Then \(\frac{x-1}{x - x^{1/2}} = \frac{b^2-1}{b(b-1)} = \frac{b+1}{b} = 1 + x^{-1/2}\).
The expression is \(\left(x^{1/3} + 1 - 1 - x^{-1/2}\right)^{10} = \left(x^{1/3} - x^{-1/2}\right)^{10}\).
General term: \(T_{r+1} = \binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r\), with power of \(x\) equal to \(\frac{10-r}{3} - \frac{r}{2}\).
Set \(\frac{10-r}{3} - \frac{r}{2} = -5 \Rightarrow 20 - 2r - 3r = -30 \Rightarrow r = 10\).
Coefficient \(= \binom{10}{10}(-1)^{10} = 1\).
Answer: (1) 1